Question: If $x \triangleright y = 8x-3y$ and $x \veebar y = 3x+y$, find $-6 \veebar (1 \triangleright -3)$.
Solution: First, find $1 \triangleright -3$ $ 1 \triangleright -3 = (8)(1)-(3)(-3)$ $ \hphantom{1 \triangleright -3} = 17$ Now, find $-6 \veebar 17$ $ -6 \veebar 17 = (3)(-6)+17$ $ \hphantom{-6 \veebar 17} = -1$.